Math 3D MT Solutions
July 6, 2016
Aaron Chen
1) Solve yy 0 + x = 0, y(3) = −4.
Solution. There’s a couple perspectives.
dy
1) Separate the equation into yy 0 = −x, which in Leibniz notation is y dx
= −x so then
ydy = −xdx −→
x2
y2
= − + C.
2
2
Because C is just a constant, we can just simplify the equation into
y 2 = −x2 + D.
Solving for the constant, we have (−4)2 = −(32 ) + D, so D = 16 + 9 = 25. Then,
p
y 2 = 25 − x2 −→ y = ± 25 − x2 .
Since the initial condition lies on x = 3 with y = −4, we will take the (-) branch! Also, we need
the square root to be non-negative, 25 − x2 ≥ 0 which tells is x2 ≤ 25, |x| ≤ 5. Hence,
p
y = − 25 − x2 , Domain of Validity: (−5, 5).
2) Substitute v = y 2 : So then
v0
2
+ x = 0 ⇐⇒ v 0 = −2x. This is separable again,
v = −x2 + C.
Plugging back in for v = y 2 , we have the same result y 2 = −x2 + C (with just a renamed constant).
The rest of the 1st solution follows from here (continue from where y 2 = −x2 + D above).
.
.
.
.
Grading Checks (needed to show work):
Separating or using substitution to start the problem and get a general solution
Solving for the constant
Solving for y explicitly and choosing the (-) branch
Finding the domain of validity.
2) General solution to y 0 + tan(x)y = cos(x).
R
tan x dx = eln | sec x| . You may have
Solution. This is an integrating
R factor problem, with R(x) = e
remembered/memorized that tan x dx = ln | sec x|. But this is why:
Z
Z
sin x
tan x dx =
dx, Use u = cos x, du = − sin xdx,
cos x
Z
−du
=
= − ln |u|, Plug back u = cos x,
u
= − ln | cos x| = ln |[cos x]−1 | = ln |1/ cos x|
1
= ln | sec x| + C
where we don’t include the constant for the integrating factor (or rather, we set C = 0 because it
doesn’t matter). Thus,
R
R(x) = e
tan x dx
= eln | sec x| = | sec x| =
1
.
| cos x|
Using this as our integrating factor, we remember that first multiply R(x) to both sides,
y0
sin x
cos x
+
y=
.
| cos x| cos x| cos x|
| cos x|
Here, we note that regardless of the branch of absolute value we take, the sign is always going to
cancel! For example, if we used | cos x| → − cos x, the minus sign just cancels everywhere. Thus,
we can just omit the absolute values,
y0
sin x
+
y = 1.
cos x cos2 x
Recalling that the integrating factor induces a product rule,
d
1
y
y = 1 =⇒
= x + C.
dx cos x
cos x
Hence, our general solution is
y(x) = (x + C) cos x.
.
.
.
.
Grading Checks:
Using integrating factor (correctly)
Finding the integral of tangent for the integrating factor
Clarifying why the sign is always positive
Correct general solution
3) General solution to y 0 =
y+xe−y/x
.
x
Solution. We notice that we should use the substitution v = y/x as a homogeneous equation because
the right hand side looks like some F (y/x). We then recall that
v = y/x ⇐⇒ y = xv, −→ y 0 = xv 0 + v.
So our ODE becomes
xv 0 + v = v + e−v ⇐⇒ xv 0 = e−v .
This is now separable, with Leibniz notation this reads as
ev
1
dx
dv
=
⇐⇒ ev dv =
.
dx
x
x
Integrating,
Z
v
e dv =
Z
dx
+ C −→ ev = ln |x| + C.
x
2
Thus, v = ln(ln |x| + C) where plugging in that v = y/x, we have that our general solution is
y(x) = x ln(ln |x| + C).
Grading Checks:
. Using the homogeneous substitution
. Correctly solving using the substitution (separation + integration)
. Plugging back in v = y/x and getting the correct general solution
4) Solve y 00 + 4y 0 + 10y = 0 with initial conditions y(0) = 3, y 0 (0) = −2.
Solution. Starting out with the auxiliary (characteristic) equation from assuming y = erx , we have
r2 + 4r + 10 = 0.
√
If you factored this: You get (r + 2)2 + 6 = 0 so then r = −2 ± i 6.
Use the quadratic formula:
√
√
√
−4 ± −24
−4 ± 16 − 40
=
= −2 ± i 6.
r=
2
2
Either way, we can now read the general solution:
√
√
y(x) = e−2x [A1 cos( 6x) + A2 sin( 6x)].
We now solve for the constants with the initial conditions. First,
y(0) = 3 : 3 = e0 [A1 cos(0) + A2 sin(0)] ⇐⇒ 3 = A1 .
To get A2 we need to differentiate (and don’t forget we can plug in for A1 too):
√
√
√
√
√
√
y 0 (x) = −2e−2x [3 cos( 6x) + A2 sin( 6x)] + e−2x [−3 6 sin( 6x) + 6A2 cos( 6x)]
so now we can use our other initial condition,
√
√
4
y 0 (0) = −2 : −2 = −2[3] + [ 6A2 ] ⇐⇒ 4 = 6A2 ⇐⇒ A2 = √ .
6
Thus, our solution after plugging in these constants is
√
√
4
y(x) = e−2x [3 cos( 6x) + √ sin( 6x)].
6
Grading Checks:
. Getting and solving the auxiliary equation
. Correct general solution from the roots of the auxiliary equation
. Solving for both of the constants using both of the initial conditions
3